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I'm trying to prove that $|\sin(x)| \le 1$, $|\cos(x)| \le 1$ and $|\sin(x)| \le |x|$ for all $x \in \mathbb{R}$ using the power series of sine and cosine :

$$\begin{align*} \sin(x) &= \sum_{k=0}^{+ \infty} (-1)^{k} \frac{x^{2k+1}}{(2k+1)!}\\ \cos(x) &= \sum_{k=0}^{+ \infty} (-1)^{k} \frac{x^{2k}}{(2k)!} \end{align*}$$

Does anyone have an idea ? I've tried to find an upper bound for the partial sums :

$$\left|\sum_{k=0}^{N} (-1)^{k} \frac{x^{2k+1}}{(2k+1)!}\right| $$

but it seems difficult.

Thanks :-)

joriki
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Louis
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1 Answers1

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Hint: Using the power series, show that $\frac{\mathrm{d}}{\mathrm{d}x}\sin(x)=\cos(x)$ and $\frac{\mathrm{d}}{\mathrm{d}x}\cos(x)=-\sin(x)$. Using those, show that $\sin^2(x)+\cos^2(x)=1$.

Added much later: This hint has been explained in more detail in this answer.

Furthermore, by the mean value theorem $\left|\frac{\sin(x)}{x}\right|=\left|\frac{\sin(x)-\sin(0)}{x-0}\right|=|\cos(\xi)|\le1$ for some $\xi$ between $0$ and $x$.

robjohn
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